Engineering Mechanics

(^506) „„„„„ A Textbook of Engineering Mechanics
Equating equations (i) and (ii),
150 g – 2 T = 150 a ...(iii)
Now consider the motion of 50 kg block, which is going upwards. A little consideration will
show that its acceleration will be (2a). We know that forces acting on it are m 2 .g = 50 g newtons
(downwards) and T newtons upwards. Therefore resultant force
= T – 50 g (upwards) ...(iv)
Since the block is moving upwards with an acceleration of (2a) therefore force acting on
this block
= 50 × 2a = 100 a ..(v)
Equating equations (iv) and (v),
T – 50 g = 100 a
Multiplying the above equation by 2,
2 T – 100 g = 200 a ...(vi)
Adding equations (iii) and (vi),
50 g = 350 a
∴^2
50 50 9.8
1.4 m/s
350 350
g
a
×
== = Ans.
and acceleration of the block B
= 2a = 2 × 1.4 = 2.8 m/s^2 Ans.
Tension in the strings
Substituting the value of a in equation (iii),
150 g – 2T = 150 × 1.4 = 210
∴ 2 T = 150g – 210 = 150 × 9.8 – 210 = 1260
or
1260
630 N
2
T== Ans.
Example 25.3. A system of masses connected by string, passing over pulleys A and B is
shown in Fig. 25.3 below :
Fig. 25.3.
Find the acceleration of the three masses, assuming weightless strings and ideal conditions
for pulleys.